SOA 289 #124

[colby2152]

I understand the SOA's solution, but I am not sure as why mine doesn't work.

, we agree there

[Gandalf]

You include the case that 3 scientists are eaten (as you should).

If three are eaten, what is the probility that exactly 2 of them exceed 8000 calories? You are using .25, and you shouldn't.

Similarly for 4,5,... scientists eaten.

[colby2152]

Originally Posted by Gandalf

You include the case that 3 scientists are eaten (as you should).

If three are eaten, what is the probility that exactly 2 of them exceed 8000 calories? You are using .25, and you shouldn't.

Similarly for 4,5,... scientists eaten.

The probability that one of them is eaten is the same as them not being eaten. Therefore, the probability that two of three are eaten is the same as none of them being eaten, or one, or three... .

Thanks for the response Gandalf, but what am I missing?

[Gandalf]

Originally Posted by colby2152

The probability that one of them is eaten is the same as them not being eaten. Therefore, the probability that two of three are eaten is the same as none of them being eaten, or one, or three... .

Thanks for the response Gandalf, but what am I missing?

That .5^3 for each case is wrong. By that logic:
p(0)+p(1)+p(2)+p(3) = 4/8. Where's the rest of the probability? exactly 2.5 exceed 8000? more than 3 of 3 exceed 8000?

The specific answer for the 3 eaten case is that p(2) = (3 choose 2)(.5^2)(.5) = .375.

The general problem with your approach is that you effectively assumed p(2) = .25 independent of how many were eaten. I don't think that approach - summing p(exactly n eaten)*p(exactly 2 over 8000 | exactly n eaten) - is feasible with a calculator, but it should work in Excel.

[colby2152]

Hmm, I did disregard the combinations since the probability was 50%. Was that my problem?

[Gandalf]

Your problem is this:

Originally Posted by colby2152

The correct formula would be

Since depends on n, you cannot convert it to a single value or take it outside the summation.

And you can't do the summation on an SOA calculator.

[jhp8]

Originally Posted by colby2152

I understand the SOA's solution, but I am not sure as why mine doesn't work.

, we agree there

Not to hi-jack the thread, but I don't really understand the SOA answer either.

"Since each time the probability of a heavy scientist is just half the probability of a success, the distribution is binomial with q = 0.6×0.5 = 0.3 and m = 8."

I understand the idea of modifying the binomial, but I am not sure why we can apply this to this problem.

[ZeroUrashima]

You just change the definition of "success".

The original N was the number of scientists eaten.
You change N to N* which is the number of scientists eaten AND above 8000 Calories.

So out of the original q=0.6, 0.3 is Above or Equal 8000 Calories, 0.3 is Below 8000 Calories. The 0.3 that is below 8000 calories would also be a a "failure" in the new distribution.

N* ~ Bino (q*=0.3,m=8)

Then Find P(N*=2)

[jhp8]

Originally Posted by ZeroUrashima

You just change the definition of "success".

The original N was the number of scientists eaten.
You change N to N* which is the number of scientists eaten AND above 8000 Calories.

So out of the original q=0.6, 0.3 is Above or Equal 8000 Calories, 0.3 is Below 8000 Calories. The 0.3 that is below 8000 calories would also be a a "failure" in the new distribution.

N* ~ Bino (q*=0.3,m=8)

Then Find P(N*=2)

Thanks....that really helped clear it up. Your explanation made me realize the piece I was missing was that point if two scientists are eaten and have > 8,000 calories, the first statement "Calculate the probability that two or more scientist are eaten" is automatically true and doesn't have to be accounted for.

[SDY2010]

I understand that q is modified to be 0.3.

The part of the solution that solves for Pr(N*=2) seems wrong. The question states that you want the probability that 2 or more scientists are eaten, not just 2.

Why would you not solve:
1-Pr(N*=0)-Pr(N*=1)
=1-(0.7^8)-[(8)*(0.3)*(0.7^7)]